449 replies. 95 posts were made by moi.Originally Posted by Paul Pless
449 replies. 95 posts were made by moi.Originally Posted by Paul Pless
"I'll tell you why [religion's] not a scam. In my opinion, all right? Tide goes in, tide goes out. Never a miscommunication. You can't explain that."Bill O'Reilly
Choosing the door that Monty didn't open is identical to having chosen both the doors you didn't choose. You win if either one has the car, regardless of the fact that Monty shows you the one that doesn't. That's 66% odds.
The table in post 24 is comically wrong, as has been pointed out multiple times in this thread. One only has to look at the first two columns to see that the number of times (rows) the contestants chooses the car is double what it should be. Yet they are treated as equal odds. Contestant chooses 1, car is behind 1 has two rows, car is behind 2 and behind 3 each have one row, same for the car being behind 2 and behind 3. Yet they are summed at the end as if that is all fine and dandy, its twice as likely the contestant chooses the car as it should be.
Ok, one last attempt for the few malcontents who refuse to accept reality. We will consider 2 different card games, the first is not the same as the Monti-hall problem, the second is equivilent, and the difference should make the answer obvious to all.
Lets find the Ace of Spades, Version 1:
The host takes a deck of cards and shuffles. The contestant picks a card at random and leaves it face down. The Host takes a card at random and leaves it face down. Both cards have a 1/52 chance of winning. The remaining 50 cards are turned up and examined. Most times, of course, the ace of spades is in the remaining 50 cards, but you do it often enough and it won't be finally. In this case, it makes no difference if the contestant switches with the host or not. The odds are 50-50.
Lets Find the Ace of Spaces, Version 2.
The host takes a deck of cards and shuffles. The contestant picks a card at random and leaves it face down. The host then takes the remaining 51 cards, goes through them and find s the ace of spades and places it face down between him and the contestant (this is equivalent to Monti opening a goat every time, as Monti's rule is the same as "he must leave the car hidden"). He then hands the 50 cards to the player to verify there is no ace of spades in it and the contestant has a choice to switch. Of course he should switch, as there is a 51/52 chance the card the host turned face down is the ace of spades and only a 1/52 chance his card is the ace of spades.
The latter game is equivalent the the problem at hand, the former game yields the result the few obstinate folks want to believe is correct.
This is how the carnies at the county fair make a living.
Not so.
We start with a bag of 3 seemingly identical objects, one of which is a valuable prize.
The probability of that bag containing the prize is 100%. The probability of any particular object in that bag, chosen at random is 33-1/3% (1/3).
The contestant selects an object from the back, at random. That object has a 33-1/3% probability of being the prize object.
Each of the remaining two objects in the bag have the same 33-1/3% probability of being the prize object, giving the bag a cumulative probability of 2/3 (66-2/3%).
Our preceptor, Monty Hall, knowing in advance, which object is the prize object and which objects are not, then draws a non-prize object from the bag and opens it, thus reducing the probability that that object was the prize to zero/zip/nada.
The bag, however, still has a 2/3 (66-2/3%) probability of containing the prize object. The 1/3 (33-1/3%) probability didn't just evaporate, it stayed with the bag, thus increasing the probability that the remaining object is the prize to 2/3 (66-2/3%).
This is Bayesian statistics at play. When the first object was selected by the contestant, it was even odds as to which object was the prize (we had no information).
Once Monty Hall reveals the non-prize object, we have introduced new information, and that changes the odds.
This is Bayesian statistics and inference at play:
Bayesian statistics is a system for describing epistemological uncertainty using the mathematical language of probability.
In the 'Bayesian paradigm,' degrees of belief in states of nature are specified; these are non-negative, and the total belief in all
states of nature is fixed to be one. Bayesian statistical methods start with existing 'prior' beliefs, and update these using data
to give 'posterior' beliefs, which may be used as the basis for inferential decisions.
http://www.scholarpedia.org/article/Bayesian_statistics
https://en.wikipedia.org/wiki/Bayesian_inference
You would not enjoy Nietzsche, sir. He is fundamentally unsound. — P.G. Wodehouse (Carry On, Jeeves)
My brain is a bit fried. I've been working all day. However, your way of counting only combines realizations when the player selects the right door at the beginning. In effect, your count cheats the player by saying "It doesn't matter." However, it means that the option to stay with the first choice is undercounted in the win columns.
"Where you live in the world should not determine whether you live in the world." - Bono
"Live in such a way that you would not be ashamed to sell your parrot to the town gossip." - Will Rogers
"Those are my principles, and if you don't like them... well, I have others." - Groucho Marx
"Where you live in the world should not determine whether you live in the world." - Bono
"Live in such a way that you would not be ashamed to sell your parrot to the town gossip." - Will Rogers
"Those are my principles, and if you don't like them... well, I have others." - Groucho Marx
"Where you live in the world should not determine whether you live in the world." - Bono
"Live in such a way that you would not be ashamed to sell your parrot to the town gossip." - Will Rogers
"Those are my principles, and if you don't like them... well, I have others." - Groucho Marx
Mythbusters episode 177
Switching won
Um... yeah.Originally Posted by L.W. Baxter
What's your point?
"I'll tell you why [religion's] not a scam. In my opinion, all right? Tide goes in, tide goes out. Never a miscommunication. You can't explain that."Bill O'Reilly
"Where you live in the world should not determine whether you live in the world." - Bono
"Live in such a way that you would not be ashamed to sell your parrot to the town gossip." - Will Rogers
"Those are my principles, and if you don't like them... well, I have others." - Groucho Marx
In a moment of clarity, I realised the key to working out the odds is that if you switch, you will always reverse your original choice.
From there - original choice has 1:3 chance of being right, so switching has a 2:3 chance.
Simples
'When I leave I don't know what I'm hoping to find. When I leave I don't know what I'm leaving behind...'
^^^^^^ this for me at #470 early this AM.
R
Sleep with one eye open.
You don't understand that you overcount correctly questing where the car is? You have to be trolling if that is the case.
The idea that a 1 in 3 possibility of winning magically becomes 1 in 2 just because someone shows you one of the wrong answers is interesting concept.
Regarding your table in post 24:
Parenthetically, for simplicity you only need to show the possibilities for 'Contestant picks Door no. 1'
For that scenario, if the car is behind door no. 1, then you can't have can't have two rows for Monty's openings of doors 2 or 3, because they are reduced to the same scenario when he asks you if you wish to switch. As others have pointed out, most recently Peb, you are erroneously doubling the 'wins' if the contestant does not switch.
That is, the tables with three scenarios for 'Contestants pick door no. 1' are correct, and your table with four scenarios is incorrect.
Moreover, the tables with three scenarios are in line with the numerous narrative distillations that do not rely on tables, as well as the computer modelings.
Now, show me where the following (numbered) logical flow is in error:
i. I have a black bag with three marbles, one is made of diamond and the others glass.
ii. You pick a marble out of the bag and hold it in your closed hand without looking at it.
iii. I ask you if you would like to exchange your one marble for the two in my bag.
iv. Would you exchange your one marble for the two in my bag?
v. Of course you would. There is a 33.3% chance that you hold the diamond marble in your hand, but a 66.6% chance that I have the diamond in the bag.
vi. Alternatively, I remove the glass marble from my bag and show it to you, then toss it aside. Because, we both well know that one of the two marbles in my bag HAS to be made of glass.
vii. I then ask you, would you like to exchange your marble for the one in my bag? Of course you would, because by showing you the glass marble I have not magically changed the probability that the bag has a 66.6% chance of holding the diamond marble.
"Where you live in the world should not determine whether you live in the world." - Bono
"Live in such a way that you would not be ashamed to sell your parrot to the town gossip." - Will Rogers
"Those are my principles, and if you don't like them... well, I have others." - Groucho Marx
I have often said that one shouldn't be allowed to graduate from high school without a solid grounding in probability and statistics.
More important than trig. More important calculus (even though you need calculus for more advance P+S)
One uses P+S every day of their life, knowingly or unknowingly -- is it safe to cross the street? Political polls, is buying a lottery ticket a good investment?
There are all sorts of things you don't consider as being probabilistic in nature.
You would not enjoy Nietzsche, sir. He is fundamentally unsound. — P.G. Wodehouse (Carry On, Jeeves)
I think the issue here is that the table treats each goat uniquely.
We’ll call them Lee and Paul.
So when the contestant chooses the correct door, you handled the outcomes of exposing Lee and exposing Paul as unique events with their own probability.
But the game deals in combinations, not permutations, so in goat form, Lee and Paul are indiscernible, and the exposing of either can be considered a singular event.
This would eliminate three rows from your table, all of which favored keeping your original choice.
"Where you live in the world should not determine whether you live in the world." - Bono
"Live in such a way that you would not be ashamed to sell your parrot to the town gossip." - Will Rogers
"Those are my principles, and if you don't like them... well, I have others." - Groucho Marx
"Where you live in the world should not determine whether you live in the world." - Bono
"Live in such a way that you would not be ashamed to sell your parrot to the town gossip." - Will Rogers
"Those are my principles, and if you don't like them... well, I have others." - Groucho Marx
Everyone is just repeating the same arguments and that includes me.
Until someone has a new idea, how about we let this go? Otherwise, we can just number the ideas we have (I think there are two and all the rest are just rephrasings of those two) and throw those numbers instead of doing all this typing.
"Where you live in the world should not determine whether you live in the world." - Bono
"Live in such a way that you would not be ashamed to sell your parrot to the town gossip." - Will Rogers
"Those are my principles, and if you don't like them... well, I have others." - Groucho Marx
meh-eh-eh-eh
put a simpler way: there are repeat numbers in the "prize behind this door" column, giving the contestant an extra win staying with each door, and an extra loss switching.
Yes, but key in this instance is that 2 times out of three when a goat is removed from play by opening a door, the other door is the car. One time out of three the other door is also a goat.
The initial choice by the contestant is meaningless other than triggering the host. It’s truely random. The host has the only real choice, but is stymied two times out of three into revealing the car by the rules of the game (he must open a door and it must be a goat). The contestant then must always choose to switch to benefit. He will only lose when the host has two goats behind two doors, which will only happen 1/3 of the time with a random shuffle.
If he doesn’t switch, he will win only 1/3 of the time because 2/3 of the time the host has one goat behind his two doors, and the car. And he must reveal the goat, and so points a finger at the car. 1/3 of the time he lies and point a finger at a goat because he has two.
Scotch and a good sleep worked for me
Last edited by Decourcy; 06-14-2022 at 12:13 AM.
To me, it makes more sense to think in terms of a game like chess. The first move isn’t designed to win, but to provoke a move that puts the pieces where you want them.